120/240V, 3 phase , 4 wires high leg delta calculations

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ghpham288

Member
Location
california
Hi All. I'm new to the forum here. I'm studying electrical engineering and having a hard time understanding the 120/240V high leg delta calculations and hoping the senior members here can help. I have the following loads below:

240 3 phase motor loads total = 38 kva
240 1 phase kitchen loads total = 21 kva
120 1 phase lighting loads total = 9 kva
120 1 phase kitchen loads total = 14 kva
120 1 phase general receptacle loads total = 9 kva

B is the stinger leg.

I've calculated the maximum feeder load to be 194 amps applying 1.25% for lighting continuous loads, and 65% for the kitchen load. Not sure if I did this correctly? Can a member please help?
 

kwired

Electron manager
Location
NE Nebraska
more details could change how this gets calculated.

I assume this is non dwelling application.

What is the largest motor? did you already factor in 125% of largest motor in the motor loads?

Which loads are able to use 220.56? Sort of assuming that is where you came up with the 65% factor you did use. Are any of the three phase motors able to fit in what is covered by 220.56?
 

ghpham288

Member
Location
california
more details could change how this gets calculated.

I assume this is non dwelling application.

What is the largest motor? did you already factor in 125% of largest motor in the motor loads?

Which loads are able to use 220.56? Sort of assuming that is where you came up with the 65% factor you did use. Are any of the three phase motors able to fit in what is covered by 220.56?

Thank you for your follow.

Yes, this is indeed a commercial space, a coffee shop indeed with some cooking equipment (ovens, blenders). All the load I have given is based on diversity factors as allowed in the code for lighting (continuous load), mechanical equipment (125% of the largest motor), and 65% diversity factor for kitchen equipment.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
I agree with kwired. There are still a lot of missing data.
I’ll try to draw a possibility.
In my opinion, 38 kVA currents are: ImotorAB, ImotorBC, ImotorCA.
kitchen 21 kVA on B and C [240 V] and kitchen120V between C and N.
Lighting and receptacle between A and N. No load between hi-leg and N.
3 currents enter point A :ImotorAB, ImotorAC, IlightAN and IreceptAN.
2 currents enter point B :ImotorBA, Ikitchen240VBC.
3 currents enter point C: ImotorCA, Ikitchen240VCB, Ikitchen120VCN.
ImotorAB=Im*(cos(fim)-jsin(fim)) ImotorBA=-ImotAB
ImotorBC=Im*(cos(-2*PI()/3-fim)-jsin(-2*PI()/3-fim))
ImotorCA=Im*(cos(-4*PI()/3-fim)-jsin(-4*PI()/3-fim))
Im=Smot/3/240V; j=sqrt(-1).
Ikitchen240V=IKT*(cos(-2*PI()/3-fik)-jsin(-2*PI()/3-fik)) IKT=21*1000/240
Ikitchen120V=Ikt*(cos(-4*PI()/3)-jsin(-4*PI()/3)) Ikt=14*0.65*1000/120
Ilight=-Ilg*(cos(-4*PI()/3)-jsin(-4*PI()/3)) Ilg=9*1.25*1000/120
Irecpt=Irc*(cos(-4*PI()/3)-jsin(-4*PI()/3)) Irc=9*1000/120
Let's say fim=ACOS(0.85) and fik [ fikitchen240]=ACOS(0.9).
lighting ,kitchen120 and recept. fi=0
Line currents:
IA=IAB+IAC
IB=IBA+IBC
IC=ICA+ICB
If I did not "cook it off" somewhere the result could be as following:
IA=91 A ,IB=175 A , IC= 126 A.
Hi_leg delta transformer flow.jpg
 

Electric-Light

Senior Member
Hi All. I'm new to the forum here. I'm studying electrical engineering and having a hard time understanding the 120/240V high leg delta calculations and hoping the senior members here can help. I have the following loads below:

240 3 phase motor loads total = 38 kva
240 1 phase kitchen loads total = 21 kva
120 1 phase lighting loads total = 9 kva
120 1 phase kitchen loads total = 14 kva
120 1 phase general receptacle loads total = 9 kva

B is the stinger leg.

I've calculated the maximum feeder load to be 194 amps applying 1.25% for lighting continuous loads, and 65% for the kitchen load. Not sure if I did this correctly? Can a member please help?

What complicates this kind of resicommercial stuff is that it could be open(fairly likely) or closed delta.
Going by the diagram posted by someone else with the A representing lower left and going clockwise.

Can we assume that 18+14 =32kVA, then 21kVA @ 240v evenly placed across C-A at unity PF and zero THD as if it's 53kw 240v single phase heater? C-A handles all the current in open delta and they would carry 221A with the 3 phase motor not running.
I'm not sure how it gets shared between the three legs for a closed system.

240v motor at unity would draw 91.4A all on its own for a closed delta.
 

kwired

Electron manager
Location
NE Nebraska
Thank you for your follow.

Yes, this is indeed a commercial space, a coffee shop indeed with some cooking equipment (ovens, blenders). All the load I have given is based on diversity factors as allowed in the code for lighting (continuous load), mechanical equipment (125% of the largest motor), and 65% diversity factor for kitchen equipment.
Is there any single phase 240 volt loads connected to the high leg is another thing to consider.
 

Electric-Light

Senior Member
Is there any single phase 240 volt loads connected to the high leg is another thing to consider.

Doing so requires double poled switches. The use of lighting apparatus with earth potential above 120v wouldn't be allowed in dwelling areas. I think its safe to assume no given the information provided.
 

kwired

Electron manager
Location
NE Nebraska
Doing so requires double poled switches. The use of lighting apparatus with earth potential above 120v wouldn't be allowed in dwelling areas. I think its safe to assume no given the information provided.
I meant is there any 240 volt loads connected A-B or B-C with B being the high leg.

A straight 240 volt rated appliance could possibly be connected to the high leg - if supplied by circuit breakers they do need to be 240 volt breakers and not 120/240 rated breakers, which are a little rare and usually expensive in comparison to 120/240 breakers, but can be done.
 

ghpham288

Member
Location
california
I meant is there any 240 volt loads connected A-B or B-C with B being the high leg.

A straight 240 volt rated appliance could possibly be connected to the high leg - if supplied by circuit breakers they do need to be 240 volt breakers and not 120/240 rated breakers, which are a little rare and usually expensive in comparison to 120/240 breakers, but can be done.

There are 240V/1ph kitchen equipment connected to A-B and B-C. I apologize but I am learning all this and 240V is a bit confusing. This is a commercial installation, lighting 120V, some general load 120V, HVAC 240V, 3 phase.
 

ghpham288

Member
Location
california
This is how I broke up the load, not at all sure if this is the correct way for the calculation?

delta2.JPG
 

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Smart $

Esteemed Member
Location
Ohio
This is how I broke up the load, not at all sure if this is the correct way for the calculation?
Here's what I get using the same loads (I think)...

Calc.png
 

kwired

Electron manager
Location
NE Nebraska
There are 240V/1ph kitchen equipment connected to A-B and B-C. I apologize but I am learning all this and 240V is a bit confusing. This is a commercial installation, lighting 120V, some general load 120V, HVAC 240V, 3 phase.
What is different is since you have limitations on what you can connect to the high leg, you won't balance the loads the same way you would on a wye system. Because of that you must know how much load is connected to where, and not just what is the total load and divide all the single phase loads as close as possible across all three phases.
 

Smart $

Esteemed Member
Location
Ohio
Here's what I get using the same loads (I think)...
Ahh... missed the general receptacle load.

Changes the results to:

Conventional 170.2A 116.1A 170.2A
Vector 183.0A 116.1A 166.3A
​ -355.9° -120.7° -235.6°
Neutral 58.3A TOTAL KVA
​ -120.0° 63.3
 

ghpham288

Member
Location
california
What is different is since you have limitations on what you can connect to the high leg, you won't balance the loads the same way you would on a wye system. Because of that you must know how much load is connected to where, and not just what is the total load and divide all the single phase loads as close as possible across all three phases.

Thank you very much.
 
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